still trying to picture the strip

nope can't imagine it

the part where it's flat would bring some trouble

because it would have to extend indefinitely and reach the other side

Actually, we don't need an infinite Möbius strip. Just do it without its boundary, but as usual.

In lectures, we're learning about curl and divergence in different coordinate systems. Quite interesting, but i'm not sure why the cross product between the gradient a vector field has anything to do with rotation

But I believe that for a generic embedding of the strip, you can extend the intervals to lines.

don't you usually parametrize the mobius strip instead of having it as a level set?

Yes, of course ...

I mean I can't really imagine a level set abruptly stopping somewhere

A level set would have to be closed, so that's why I said infinite ....

and in my imagination the flat bit would cause problem

But certainly hypersurfaces needn't be orientable when they're not compact without boundary.

I don't know what flat bit you're talking about. The usual parametrization isn't flat anywhere.

The interval keeps rotating constantly ....

i.e. parallel to the xy-plane

The model I usually write down is negatively curved pretty much everywhere, I'm pretty sure.

@TedShifrin, which of your lectures do you suggest for me to watch before tackling the implicit function theorem, and the inverse one?

do you actually have a smooth function with a level set homeomorphic to the mobius strip though

I can't tell

sorry for being distracting

because I don't see why my construction of the orientation won't work

given any closed set in $\Bbb R^n$ there's a smooth function with that as level set.

hmm...

but if I stipulate that the grad has to not vanish on the set then there might not be one, right

yes

@topologicalmagician: If you want to understand proofs, you need a lot of derivative background. To understand examples, you can just watch those, I guess.

because I can just specify the orientation by the grad, right

yes

great thanks :)

I decided to do my taxes since that sounds like less of a pain than working out some signs I need to do

who invented exterior derivative?

(and in which paper?)

also Arnold's book seems to be behind paywalls?

A lot of the exterior calculus was developed by Elie Cartan. I should check if it was around (other than informally) before him. I should know this.

thanks

@LeakyNun most books are behind paywalls

you can find it in the standard places, though

Are libraries still standard places?

Generically, yes

@LeakyNun cartan

Ah, @Leaky, Wiki says i was right and gives an 1899 paper of Cartan. Too bad I no longer have the 4 volumes of the complete works of E. Cartan.

1899

thanks all

some paper about pfaffian eqs

@TedShifrin Yes, among others.

you matched wiki perfectly @ÉricoMeloSilva

I actually looked at that paper years ago.

impressive

Of course, Leaky could have wikied himself.

I prefer paper copies and so that's my first place to look.

well I don't know if that was the first paper...

In the early days, he wrote $d$ and $\delta$ as "independent" exterior derivatives.

Chern was the first to give a proof of Gauss-Bonnet for surfaces using forms, I believe, and then, of course, he generalized it completely ...

i read some book cartan wrote where it described the developments

idr what it was called

Elie or Henri?

Prove that $\displaystyle \int_0^\infty \frac{\mathrm dx}{1+x^2} = \int_{-1}^1 \sqrt{1-x^2} \ \mathrm dx$ using only algebraic (multivariate) substitutions, linearity of integration, partition of area under integration, and Stokes theorem [I don't know if it is possible]

no

there is an "elementary" (in the above sense) proof that zeta(2) = pi^2/6

and it very much seems like magic

nobody would come up with those substitutions

One such proof is an exercise in my book. I stole it from Simmons's calculus book.

Well, seeing $1-xy$ it isn't so hard to come up with a $45^\circ$ rotation change of variables.

I suppose

Think normal forms for conics.

@TedShifrin papa

i read henris book on forms as well ofc

@Eric: Yeah, I have read parts of that book (in French) ... Maybe I can resurrect the title by looking.

A teacher of mine recommended a book called calculus by james stewart. I really enjoyed it when I borrowed the book, but its a shame he doesn't discuss manifolds

I think it was Leçons sur les Invariant Intégraux.

Your teacher made a terrible suggestion, unfortunately.

@topologicalmagician: Most of us aren't so fond of that book. He ripped off Edwards & Penney quite a bit.

That's probably the worst calculus book I know.

Their second edition was a fabulous book.

I don't know any good calculus books.

I don't know any good [-] books.

I resemble that remark, @MikeM.

i have also read the letters exchanged between elie and einstein @Ted

OK, I walk it back.

lots of great history there

Yeah, I think I saw a little of that, Eric.

I might know some good books. I just haven't read them.

@MikeMiller ik some but they’re not math

I definitely have some favorite math books. And I won't even be egotistical :P

how would you explain cohomology to your son?

oh

I would put him out of his misery before his brain developed enough to comprehend the human condition.

Oy.

Maybe on that note I should go pack for my trip.

My bad.

Enjoy the trip, though.

@TedShifrin Have a safe trip!

I'm not leaving yet ... :) Just avoiding this particular conversation.

@Leaky I'd give the answer that I always want to give when someone says "ELI5"

"You'll learn this when you get older"

(Assuming said son were young)

wow Pfaff goes way back

@Daminark yep, should have definitely written that on my form

I mean maybe you can make your son near college age. In which case, I dunno my instinct would be to explain De Rham as something like "So you know how you can't extend $\frac{1}{x}$ to a continuous function on all of $\mathbb{R}$? Cohomology measures these kinds of obstructions to defining certain types of functions"

That's at least my current internal image of cohomology, perhaps replacing "functions" with "forms". No idea if that's good or not

wait cohomology tells you about 1/x?

Why not homology/cohomology indicate different-dimensional holes ... ?

No I just mean, closed forms mod exact forms in my mind sorta tells you how far you are away from being able to define the primitive of a form.

And I have that image for homology but not cohomology

well, you detect the holes by cohomology :P

think about $(-y\,dx+x\,dy)/(x^2+y^2)$ and the analogous Gauss's law 2-form in $\Bbb R^3-\{0\}$.

Hmm, so would that be something like, $\frac{-ydx + xdy - zdx + xdz - zdy + ydz}{x^2 + y^2 + z^2}$?

$2$-form

Right I'm blind

Think inverse square force field.

Uh, at this point I'm kinda just trying to do a symbolic analogy more than anything, my physics is practically nonexistent

wait a minute... did Cartan define exterior derivative and then right afterwards state that it commutes with pullback

so somehow that is the most important property?

Well, forget physics. Write down a vector field $F$ that is rotationally symmetric, points radially, and has $\|F\| = 1/r^2$ at distance $r$. Turn it into a $2$-form.

None of us here has the original source handy.

(also is he using $\varpi$ (pi) for $\overline{\omega}$ (omega))?

Yes, then, is the answer to your question. They typeset it that way. I'm sure he didn't intend it.

also I'm not sure why his product is commutative...

The whole point of forms is changes of coordinates. That's what the change of variables theorem teaches you, Leaky.

Forms are what it makes sense to integrate.

What's confusing in classic literature is the confusion with exterior product (not written $\wedge$) and symmetric tensor product (not written $\otimes$). Ha!

then why is omega 1 omega 1 = 0?

Wedges omitted.

Read what I just typed ^^^

so.. it's both at the same time, you mean?

No, you have to figure out from context what they're doing.

Even in modern days, people write the metric as $g_{ij}dx^i\,dx^j$ without putting in the tensor sign.

in no context would (ω1 + ω2)^2 be 2ω1ω2, is what I mean

Yeah, that's wedging the sum with itself.

that would give 0?

Hey I'm back! So $(x,y,z)$ is the obvious candidate, and that converts to a 2-form as $\frac{x dy \wedge dz - y dx\wedge dz + z dx \wedge dy}{x^2 + y^2 + z^2}$

if I have a function f:{x0} -> R is it strictly monotone increasing?

@famesyasd depends on definition of "strictly monotone increasing"

but it is yes in my definition

forall x1,x2 in dom(f) if x1 < x2 then f(x1) < f(x2)

Hmm, let me try to figure out what Cartan is doing.

then yes

@Ted ok thanks

I need timeout to get a martini first.

@TedShifrin ik folk who stick a dot for symmetric prod

but i just supress

So what is Cartan doing there, Eric?

iunno i was tabbed out im just passing through lmao

also Cartans says $\omega'' = \omega \omega'$ and $\omega''' = \frac12 \omega'^2$ and $\omega'''' = \omega \omega'''$

I'm completely lost

Demonark: Except for your power on the denominator, you're right. Try taking $d$ to get the power right. :P

Leaky, if you read what he said, he's considering the case where $\omega_j$ are forms of even degree. He said it would be $0$ if they're odd. He's also thinking of each one as a monomial, not a "polynomial" in the exterior algebra. That's why $\omega_j\wedge\omega_j = 0$.

For that stuff with primes, you'll have to give me the section.

oh ok...

it's page 15

reading papers is really a whole nother business...

thanks for your patience

@Leaky: I don't have the book, so you'll have to show me.

Oh, $\omega'' = \omega\wedge d\omega$, right?

And he's defining $\omega'''$ to be $\frac12 d\omega\wedge d\omega$.

but why?

This relates to integrability conditions for things like contact forms.

oh ok

You look at $\omega\wedge (d\omega)^k$, etc.

so it isn't related to the stuff I'm doing?

No, not in the least.

ok

oh hell, now @Erico is back again.

i aint back

Oh, whew.

just opened my laptop lol

Why does it autoload here?

Cuz you stay in the room instead of logging out?

i didnt shut down my browser so this tab was still open

AH.

i prob wont be around chat much till after grad visits finish up

Well, keep me posted!

indeed

6 visits :( so much flying

me no like

Happy travels and visits.

tyty

You couldn't combine any of 'em?

the UCB and stanfo visits i managed to combine thankfully

they're the same weekend and they have a bus taking ppl from stanford to UCB lmao

That's not too surprising. I'm sure they coordinated on porpoise.

Saves them beaucoup de money, too.

yeah most definitely

they wont have to pay for all those flights

I'm out of here for now. Safe travels and hunting!

ty tchau tchau

@Albas There is a famous quote from Arnold about contact geometry and how it's required to understand thermodynamics. Arnold is also famous for making heavy-handed comments, or at least in my opinion. The amount of modern contact topology that makes an appearance in thermodynamics is very small. The amount of contact geometry I have personally seen applied to thermodynamics is also quite small, and could be summarized as "you can phrase this alternatively with contact geometry".

@Albas It doesn't seem particularly enlightening, other than being succinct, and nice looking. That being said, I definitely agree that it is super cool to see some mathematical formalisms put forth for topics in physics, so in light of doing so, I will provide some good references.

@Albas There is firstly this paper: sci.sdsu.edu/~salamon/MathThermoStates.pdf Then this one: kurims.kyoto-u.ac.jp/~kyodo/kokyuroku/contents/pdf/1142-13.pdf And then there is Geiges's outline on contact manifolds which gives general theory as well as applications: sciencedirect.com/science/article/pii/S0723086901800141

@Albas if at any time you have any questions about contact topology, feel free to message me here or whatever. I enjoy talking about it. It also helps me learn it better, as someone who is relatively new to research in the field.

hi, I need to find the midpoint of a hyperplane given inequality constraints on the coordinates

i have a post on math.stackexchange.com/questions/3114641/… with 4 upvotes but no biters

any hints would be appreciated

Hm, @TedShifrin, I'm ending up in the same place. So, yes, agreed, my matrix is $A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$. Now $A \cdot \begin{bmatrix} 1\\ 0 \end{bmatrix} = \begin{bmatrix} 0\\ 0 \end{bmatrix}$, and $A \cdot \begin{bmatrix} 0\\ 1 \end{bmatrix} = \begin{bmatrix} 1\\ 0 \end{bmatrix}$. That is, $R(A) = \mathbb{C} \times \{0\}$.

And $N(A) = span\{\begin{bmatrix} 1\\ 0 \end{bmatrix}\} = \mathbb{C} \times \{0\}$. So $R(A) + N(A) = \mathbb{C} \times \{0\} \ne \mathbb{C}^2$. Why am I crazy?

To be a little more pedantic -- $R(A) = span\{\begin{bmatrix} 1\\ 0 \end{bmatrix}\} = \mathbb{C} \times \{0\}$.

What's wrong with $R(A) + N(A) \neq \mathbb{C}^2$?

@Pig there's a claim that states that if $A$ is $n \times n$ (over $\mathbb{C}$), then $\mathbb{C}^n = R(A) \oplus N(A)$, and I claim this claim is wrong

btw, @TedShifrin, I was using those as column vectors, not row vector. All kosher.

oh yeah, that's false for sure

how do you know?

(other than my counter example)

yours is pretty much how all such examples came about

as long as $R(A) \cap N(A) \neq \{0\}$ the statement can't be true

well the sum doesn't even give you everything you want

and constructing $A$ where they intersect - nilpotent jordan block is pretty much the "main" thing that happens

forget about a direct sum

as far as i remember a necessary condition is $R(A^k) = R(A^{k+1})$, in which case $\mathbb{C}^n = R(A^k) + N(A^k)$

where the necessary condition obviously doesn't hold in my example

@anakhro These are wonderful

I feel like this is a dumb question, but is $\lim_{h \to 0} \frac{f(x + h) - f(x)}{h} = \pm \lim_{h \to 0} \frac{f(x - h) - f(x)}{h}$? Assuming $f : \mathbb{R} \to \mathbb{R}^n$ is differentiable.

The systole of a compact metric space $X$ is defined to be the infimum of the length of a noncontractible loop in $X$ (i.e. a loop that cannot be contracted to a point in the ambient space $X$). We denote the systole of $X$ by $sys(X)$. Let $M$ be a compact Riemannian 3-manifold and suppose that $M$ has a closed embedded nonorientable surface (i.e. a sphere with $h$ cross caps). Does this imply that $sys(M)>0$?

@WilliamOliver $\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to 0}\frac{f(x-h)-f(x)}{-h}=-\lim_{h\to 0}\frac{f(x-h)-f(x)}{h}$

@Derso thats what I thought, thanks

Hey

A linear transformation is the same that linear operator?

I feel like I might have just made a stupid mistake when deriving (2), but I've been over it a few times, and somehow wolfram alpha agrees.

@JoeShmo: Before I leave town for the weekend — the issue is notation. That problem is using the same notation I use in my books. $R(A)$ is the row space of $A$, not the range (image = column space) of $A$. Note that the range and nullspace live in different vector spaces, so it makes no sense to take their direct sum.

Oh wait I think I just realized the answer to my question! I am pretty sure the limit $\frac{c'}{||c'||}$ for an arbitrary curve $c$ doesn't exist, because as $h$ approaches 0 from one side, the vector is one direction, but as it approaches from the other side, the vector is the other direction! So the product law doesn't apply.

@TedShifrin ohh... valid point. I should have clarified with him what $R(A)$ is.. I took it as the range range (image = column space) obviously

And the range and null space indeed live in different vector spaces, but if $A: \mathbb{C}^n \rightarrow \mathbb{C}^n$, and $Range(A) \cap Nullspace(A) = \{0\}$, then in fact it's true that $\mathbb{C}^n = Range(A) \oplus Nullspace(A)$

I'm in LA for the day actually, drove down from SF all of yesterday via route 1. Almost went down to SD, but don't have time

@JoeShmo They still live in different places. There you're just making a dimension argument that any two complementary subspaces add up to the whole thing. It's not right, however, because $R(A)$ is NOT a subspace of the same space ...

Route 1 is a slow drive. It just recently reopened. It's OK. I'm leaving in an hour for a long weekend in Palm Springs.

hi italic @Alessandro

Hi @Ted

At any rate, @JoeShmo, you truly have to say isomorphic there, if you're going to write that. Equals is impossible, as I explained.

The point is that $A$ maps $R(A)$ isomorphically to $C(A)$ (rows to columns).

if $A$ is idempotent then it is a equal sign

but at any rate he needs to provide source

instead of having us figure out what his source meant

I'll look at it, I thought I could prove actual equality in that case (which I don't contend in the general case anyway)

route 1 is 2 hours slower that route 5, but the scenes are breathtaking

why do they still live in different places? $A$ maps $\mathbb{C}^n$ to itself

hello

hi

hello

@RajendraBalliwal Try to negate this statement: "if Jack is tall then Jack is strong"

first thanks for help

we'll see if this helps....

i would say it as " - ( tall -> strong ) "

No, don't use math, use your brain

Say it in words

Assume you know jack

if jack is strong then jack is tall

is that correct ?

that isn't the opposite

yeah its true when the above is false

ok, suppose I say "If Jack is tall then Jack is strong", but you know I'm lying. Tell me about jack.

Is jack weak or strong?

this means that jack is tall but not strong

yes!

so , how does it helps ?

the opposite of "If jack is tall then jack is strong" is "Jack is tall and not strong"

that was step 1

now step 2

tell me the negation of "If a man is tall, then a man is strong"

Guys is the same linear transformation=linear operator?

according to above analogy it comes as , if a man is tall then he is not strong

but I think you know, "If a man is tall, then he is strong" is false, and I think you know "If a man is tall then he is not strong" is also false

i am really finding it hard to tell negation of " impications "

so they can't be opposite

stop hiding behind math symbols and think

Why don't make a truth table

yeah sure

because you should know the opposite of "If a man is tall, then he is strong" without doing any math

use your brain, your experience, your life

@DanielML usually linear operator means exactly linear transformation from V to V (same vector space)

@RajendraBalliwal Are you going afk?

@DanielV yeah iam trying

Ok, suppose a stranger walked up to your friend and said "If a man is tall, then a man isi strong". Do you believe him?

i have made simple negation of and , or but of not implication thus searching , please give me a moment and dont go away as i have a lot of questions

ok, I see you are determined to hide behind math symbols and not answer my question

I give up

no

no

You got 10 seconds to answer my last question or I'm leaving

ok well you are busy see you

i dont belive

as a man can not be tall but strong

in that case the statement is false

no no , its false because it fails to justify the condition

when a men is tall but not strong

and this can happen

@DanielV negation is a men is tall but not strong

Does a distortion of gridlines imply curvature

@DanielV i think its not implication

its "or"

thus the neagation is " a men is not tall and he is not strong

i think i got it right this time

I think all these geodesics imply that the underlying manifold is curved

Hello. I am wondering if someone would help me in understanding the wedge sum of a collection of topological spaces. Let $\{X_i\}_{i \in I}$ be a collection of topological spaces. Then $\amalg_{i \in I} X_i := \cup_{i \in I} \{(x,i) \mid x \in X_i\}$ is the disjoint union of the topological spaces. Let $x_i \in X_i$.

From what I understand the wedge is defined as $\vee_{i \in I} X_i := \amalg_{i \in I} X_i/ \sim$ with the equivalence relation $\sim$ defined such that all points $(x_i,i)$ are equivalent/identified (i.e.,they all lie in the same equivalence class).

Does this sound right?

yes, that's correct

How do the submodules of $M \otimes_R N$ compare to that of $M$ and $N$?

can anyone help me understand universal and existential elimination/introduction in first order logic?

And why to call operator when u can say transformation is the same thing lol

@RajendraBalliwal The negation of a statement $p \implies q$ is $p and not q$

For instance, consider: If I like all cars then I like red cars. The negation of that is I like all cars and I do not like red cars. Think about it. If I like all cars then I like red cars is true then I like all cars and I do not like red cars is false.

how do you prove that not forall x, P(x) is the same as exists x, not P(x)?

what does $\Bbb R \times \Bbb C$ mean

Question

I need to visualize the statements. I mean why to make that distinction. For example I have:

let (V,<>) a vector space with inner product of finite dimension and the same statement but why to say finite dimension and dimension n

what is a smooth coordinate chart on a manifold

what is the CW complex

what is the systole of an american football?